H****h
发帖数: 1037 | 1
Since |a+b|\le |a|+|b|\le either 2|a| or 2|b|,
so |a+b|^r\le (2|a|)^r+(2|b|)^r=2^{r+1}(|a|^r+|b|^r).
You may choose C_r=2^{r+1}.
Of course this C_r is not the smallest. | n*******l
发帖数: 2911 | 2 f(x) = |x|^r is convex for r>=1.
So f( a/2 + b/2 ) < = f(a)/2 + f(b)/2.
Simplify the above inequality, we can get the desired inequality. |
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